Nice work. It looks like this part is off
opulence2001 Wrote:
S(most) -> C(all)--> R(all)
which in inverse work out as:
R(some) --> C(some) --> S(all)
The inverse of S-(most)-> C is actually C-(some)-> S . In short,
most and
some reversed become
some. BTW, it might be easier for you to simply write the "some" or "most" above the arrow if you use this sort of notation. There's only one modifier per relationship.
I agree that the "not all skilled artists are famous" is a distraction.
The reasoning is basically that you need S -(most)-> R (because S-(most)-> C-(all)-> R) and S -(most)-> F to ensure an overlap between R and F.
I believe you've incorrectly written that S-(all)-> F, all we know is that S-(most)-> F.
The key is to know a few some and most rules:
some + some = ?
some + most = ?
most + most = overlap
With this sort of question, it's totally possible that I messed up, so fire away!